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Prove $\|u_k\|$ converges to $\|u\|$ if $u_k$ converges to $u$. Assume $(u_k)_{k=1}^{\infty} \subset \mathbb{R}^n$ and $u \in \mathbb{R}^n$.

To solve this problem, I remembered a trick from Algebra:

\begin{align} \sqrt{x}- \sqrt{y} &=\sqrt{x} - \sqrt{y}\frac{\sqrt{x}+ \sqrt{y}}{\sqrt{x}+ \sqrt{y}}\\&=\frac{x-y}{\sqrt{x}-\sqrt{y}} \end{align}

Similarly,

\begin{align} \left| \|u_k\|-\|u\|\right| &=\left| \sqrt{\sum_{i=1}^n p_i(u_k)^2}- \sqrt{\sum_{i=1}^n p_i(u)^2} \right| \\ &=\left| \frac{\sum_{i=1}^n p_i(u_k)^2-\sum_{i=1}^n p_i(u)^2}{\sqrt{\sum_{i=1}^n p_i(u_k)^2} -\sqrt{\sum_{i=1}^n p_i(u)^2}} \right| \\ &\leq \frac{\left|\sum_{i=1}^n p_i(u_k)^2-\sum_{i=1}^n p_i(u)^2 \right|}{\sqrt{\sum_{i=1}^n p_i(u_k)^2} -\sqrt{\sum_{i=1}^n p_i(u)^2}} \\ &\leq \frac{\left|\sum_{i=1}^n p_i(u_k-u)^2 \right|}{\sqrt{\sum_{i=1}^n p_i(u_k)^2} -\sqrt{\sum_{i=1}^n p_i(u)^2}} \end{align}

Now I would like to use the another algebra trick where if $x > z$ implies

$$\frac1{x+z} \leq \frac1{y+z}.$$

But I am stuck at this point. I know I need to exploit the fact that $u_k$ converges to $u$. Would appreciate all / any help from community.

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For all norms, following inequality holds: $$\vert \Vert x \Vert - \Vert y \Vert \vert \le \Vert x-y\Vert$$

which is enough to prove what you want.

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Hint:

Just use the reverse triangle inequality which holds true for any norm:

$$\left| \, ||u_k|| - ||u||\, \right| \leq ||u_k - u||$$

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